其实就是在扫描线填充三角形的基础上,对三个顶点的xy坐标计算三角形重心坐标,然后在填充时,原来三角形重心坐标填充是遍历整个屏幕像素,然后判断是否在在三角形内,变成根据直线方程计算x起点与y起点,然后将需要求重心坐标的点zx,zy,从遍历屏幕像素的点xy,换成扫描线起始点xy
for gao in range(280):
xz1 = (jz - (gao+by)) / zk
sx = abs(xz1)
xy1 = (jy - (gao+by)) / yk
ex = abs(xy1)+2#加2是为了填补空白
xcishu=ex-sx
#print("gao==",gao,"第一次循环",int(sx),int(ex))
for i in range(int(xcishu)):
#print("i==",i,"第二次循环",int(sx),int(ex))
zx = int(sx)+i
zy = gao+by
将原来的
#pygame.Surface.set_at(screen, (int(sx)+i , by+gao), (255, 0, 0))
换成
if (0 < zhongxina < 1 and 0 < zhongxinb < 1 and 0 < zhongxinc < 1): pygame.Surface.set_at(screen, (int(sx)+i, int(gao+by)),(int(255 * zhongxina), int(255 * zhongxinb), int(255 * zhongxinc)))
加by是因为
for gao in range(280):
280是三角形上顶点与下边之间的高度,还需要加上上顶点的y才能填充580这个高度的三角形,不然出现两个三角形
import pygame
import sys
from pygame import gfxdraw
(width, height) = (600, 600)
pygame.init()
screen = pygame.display.set_mode((width, height))
image = pygame.image.load("1.png").convert_alpha()
masked_result = image.copy()
white_color = (0, 0, 0)
polygon =[(0, 0), (800, 600), (0, 600)]
#(260,0),(200,0),(200,130),(260,130)
#[(0.0, 307.6923076923077), (